Question

这是编码chucknorris游戏的一部分，当我们想要开始一个字符串，然后将其转换为“一元”时，一个只有0和空格的代码。这是我的代码，我的问题是没有任何东西可以解决我的问题。例如： CC =＆gt; 10000111000011：

0 0 (one 1)
00 0000 (four 0)
0 000 (three 1)
00 0000 (four 0)
0 00 (two 1)

CC给出：0 0 00 0000 0 000 00 0000 0 00

import sys
import math
# Auto-generated code below aims at helping you parse
# the standard input according to the problem statement.*

message = input()
# message to binary (01001)
bimsg = str([ bin(ord(ch))[2:].zfill(8) for ch in message ])[2:-2]

# function
to_chuck = ''
for n in bimsg:
    index = bimsg.index(n)
    if index >= 1 and bimsg[index] == bimsg[index-1]:
        to_chuck+='0'
    else: 
        if n == 1:
            to_chuck+='0 0'
        elif n == 0:
            to_chuck+='00 0'

print(to_chuck)

Answer 1

您的代码中有4个错误。

将输入字符串转换为二进制文件的方式不正确。您将每个字符填充为8位而不是7位，并且由于您在列表上调用str，因此具有多个字符的字符串将在字符之间显示逗号。正确的转变是
```
bimsg = ''.join(format(ord(x), 'b').zfill(7) for x in message)
```
index = bimsg.index(n)将始终为您提供输入中n的第一次出现的索引。如果您的输入为1010，则bimsg.index('1')将始终返回0，bimsg.index('0')将始终返回1.这不是您想要的。

要获得正确的索引，请使用enumerate函数：
```
to_chuck = ''
for index, n in enumerate(bimsg):
    if index >= 1 and bimsg[index] == bimsg[index-1]:
        to_chuck+='0'
    else: 
        if n == 1:
            to_chuck+='0 0'
        elif n == 0:
            to_chuck+='00 0'
```

由于bimsg是一个字符串，n也是一个字符串。当您执行if n == 1或if n == 0时，您需要将字符串与数字进行比较 - 这将始终评估为False。您必须将n与字符串进行比较：

to_chuck = ''
for index, n in enumerate(bimsg):  # use enumerate here
    if index >= 1 and bimsg[index] == bimsg[index-1]:
        to_chuck+='0'
    else: 
        if n == '1':
            to_chuck+='0 0'
        elif n == '0':
            to_chuck+='00 0'

您的代码不会在这里的0和0之间放置任何空格：

if n == '1':  # changed 1 to '1' here
    to_chuck+='0 0'
elif n == '0':  # changed 0 to '0' here
    to_chuck+='00 0'

你添加的零只是在to_chuck的末尾与零一起融合，给你错误的输出。要解决此问题，只需添加一个空格：

to_chuck = ''
for index, n in enumerate(bimsg):
    if index >= 1 and bimsg[index] == bimsg[index-1]:
        to_chuck+='0'
    else: 
        if n == '1':
            to_chuck+=' 0 0'  # added leading space here
        elif n == '0':
            to_chuck+=' 00 0'  # added leading space here
to_chuck = to_chuck[1:]  # get rid of the leading space

现在我们得到正确的输出：

print(to_chuck)
# 0 0 00 0000 0 000 00 0000 0 00

Answer 2

def firstBlock(x):
    if x == "1":
        return "0"
    return "00"

def getBlock(x, counter):
    return firstBlock(x) + " " + ("0" * counter)

message = input()
result = ""

binaryString = ""
for x in message:
    binaryChar = format(ord(x), 'b')
    paddedChar = binaryChar.zfill(7)
    binaryString += paddedChar

counter = 1
currentChar = binaryString[0]

for x in binaryString[1:]:
    if x == currentChar:
        counter += 1
        continue

    result += getBlock(currentChar, counter) + " "
    counter = 1
    currentChar = x

result += getBlock(currentChar, counter)
print(result)

从二进制到一元（Codingame.com）

2 个答案: